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NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity - Welcome to the updated current electricity class 12 exercise solutions for one of the easy and intriguing chapters in the Class 12 syllabus. On this page, you will discover comprehensive NCERT solutions spanning from question 3.1 to 3.23. Questions 3.1 to 3.14 are part of the exercise section, while the rest are from the additional exercise. These class 12 physics chapter 3 ncert solutions have been meticulously crafted by subject experts, aligning with the latest syllabus and offering detailed explanations.
Physics Class 12 chapter 3 NCERT solutions are related to Ohm’s law and Kirchoff's law which will help in solving circuit related problems in exams. The NCERT Solutions for Class 12 Physics Chapter 3 also explains problems related to heat developed in a resistor. The basics of these chapters are covered in NCERT of high school classes also.
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Current is the flow of electric charge through a given area. If we want to utilize the charge flow, we must have a closed path. NCERT Solutions For Class 12 Physics chapter 3 explains questions about the factors affecting the current flow, how current flows in a conductor, basic laws related to current electricity and circuit analysis and a few measuring devices. NCERT Solutions for Class 12 contain the answer to each question given in NCERT Class 12 Physics book.
Free download class 12 physics chapter 3 exercise solutions PDF for CBSE exam.
Given, the emf of battery, E = 12 V
Internal resistance of battery, r = 0.4 Ohm
Let I be the maximum current drawn from the battery.
We know, according to Ohm's law
E = Ir
I = E/r = 12/0.4 =30 A
Hence the maximum current drawn from the battery is 30 A.
Given, The emf of the battery, E = 10 V
The internal resistance of the battery, r = 3 Ohm
Current in the circuit, I = 0.5 A
Let R be the resistance of the resistor.
Therefore, according to Ohm's law:
E = IR' = I(R + r)
10 = 0.5(R + 3)
R = 1 Ohm
Also,
V = IR (Across the resistor)
= 0.5 x 17 = 8.5 V
Hence, terminal voltage across the resistor = 8.5 V
We know that when resistors are combined in series, the effective resistance is the sum of that resistance.
Hence, total resistance of the three resistance combination = 1 + 2 + 3 = 6 Ohm
Answer:
Since the resistances are in series, the current through each one of them will be equal to the current through the circuit but voltage/ potential drop will be different.
Total resistance, R = 6 Ω
Emf, V = 12 V
According to Ohm's law:
V = IR
12 = I x 6
I = 12/6 = 2 A
Now, using the same relation, voltage through resistors:
1 Ω : V(1) = 2 x 1 = 2V
2 Ω : V(2) = 2 x 2 = 4V
3 Ω : V(3) = 2 x 3 = 6V
(Note: V(1) + V(2) + V(3) = 2 + 4 + 6 = 12 V )
We know that when resistances are in parallel combination, the total resistance R is given by:
Therefore, total resistance of the given three resistances in parallel combination is
Hence, the total resistance is 1.05 Ω
Since the resistances are in parallel, the voltage across each one of them will be equal.
Emf, V = 20 V
According to Ohm's law:
V = IR I = V/R
Therefore, current across each one of them is:
2 Ω : I = 20/2 = 10 A
4 Ω : I = 20/4 = 5 A
5 Ω : I = 20/5 = 4 A
3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.7 x 10^{-4 o}C^{-1}
Given,
temperature coefficient of filament,
α = 1.7 x 10^{-4 o}C^{-1}
;
Let T_{2} be the temperature of element, R_{2}=117 Ω
(Positive alpha means that the resistance increases with temperature. Hence we can deduce that T_{2} will be greater than T_{1} )
We know,
Hence, the temperature of the element is 1027 °C.
, where is the resistivity of the material
Hence, the resistivity of the material of wire is
3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Given,
;
;
We know,
2.7 = 2.1[1 + (100 - 27.5) ]
= (2.7 - 2.1) / 2.1(100 -27.5)
= 0.0039
Hence, the temperature coefficient of silver wire is 0.0039
For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.
V = 230 V
and
Using Ohm's law:
and
Let be the steady temperature of the heating element.
We know,
That is
230/2.8 = 230/3.2[1 + ( ) ( T _{ 2 } - 27) ]
T _{ 2 } = (840.5 + 27) °C
Hence, steady temperature of the element is 867.5 °C .
3.9 Determine the current in each branch of the network shown in Fig. 3.30:
Answer:
Let current in the circuit is distributed like
where I_{1}, I_{2}, and I_{3} are the different current through shown branches.
Now, applying KVL in Loop
Also, we have
so putting it in kvl equation
.................................(1)
Now let's apply kvl in the loop involving I1 I2 AND I3
.................................(2)
now, the third equation of KVL
..............................(3)
Now we have 3 equation and 3 variable, on solving we get
Now the total current
Balance point from the end A, l _{ 1 } = 39.5 cm
Resistance of Y = 12.5 Ω
We know, for a meter bridge, balance condition is:
The connections between resistors in a Wheatstone or meter bridge made of thick copper strips to minimise the resistance of the connection which is not accounted for in the bridge formula.
(b) Determine the balance point of the bridge above if X and Y are interchanged.
If X and Y are interchanged.
Then, X = 12.5 Ω , Y = 8.2 Ω
We know, for a meter bridge, balance condition is:
if the galvanometer and cell are interchanged the galvanometer will show no current and hence no deflection.
Emf of battery, E = 8 V
Internal resistance of battery, r = 0.5 Ω
Supply Voltage, V = 120 V
The resistance of the resistor, R = 15.5 Ω
Let V' be the effective voltage in the circuit.
Now, V' = V - E
V' = 120 - 8 = 112 V
Now, current flowing in the circuit is:
I = V' / (R + r)
Now, using Ohm ’s Law:
Voltage across resistor R is v = IR
V = 7 x 15.5 = 108.5 V
Now, the voltage supplied, V = Terminal voltage of battery + V
Terminal voltage of battery = 120 -108.5 = 11.5 V
The purpose of having a series resistor is to limit the current drawn from the supply.
Let E_{2} be the voltage in the second case.
Now, the balance condition is given by :
Therefore, the emf of the second cell = 2.25 V
We know,
v_{d} :drift Velocity = length of wire(l) / time taken to cover
by substituting the given values
t = 2.7 x 10^{4} s
Therefore, the time required by an electron to drift from one end of a wire to its other end is s.
Given, the surface charge density of earth
=
Current over the entire globe = 1800 A
Radius of earth, r = 6.37 x 10^{6} m
The surface area of earth A =
= =
Now, charge on the earth surface,
Therefore,
Let the time taken to neutralize earth surface be t
Current I = q / t
t = 282.78 s.
Therefore, time take to neutralize the Earth's surface is 282.78 s
Given,
There are 6 secondary cells.
Emf of each cell, E = 2 V (In series)
The internal resistance of each cell, r = 0.015 Ω (In series)
And the resistance of the resistor, R = 8.5 Ω
Let I be the current drawn in the circuit.
I = 1.4 A
Hence current drawn from the supply is 1.4 A
Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V
Given,
Emf , E = 1.9 V
Internal resistance, r =380 Ω
The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A
The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.
We know,
R = Ρ l / A
The wires have the same resistance and also are of the same length.
Hence,
Now, mass = Density x Volume = Density x Area x length
Taking the ratio of their masses for the same length
Hence,
Therefore, for the same resistance and length, the aluminium wire is lighter.
Since aluminium wire is lighter, it is used as power cables.
3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?
Current A | Voltage V | Current A | Voltage V |
---|---|---|---|
0.2 | 3.94 | 3.0 | 59.2 |
0.4 | 7.87 | 4.0 | 78.8 |
0.6 | 11.8 | 5.0 | 98.6 |
0.8 | 15.7 | 6.0 | 118.5 |
1.0 | 19.7 | 7.0 | 138.2 |
2.0 | 39.4 | 8.0 | 158.0 |
The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.
Hence the resistor made of alloy manganin follows Ohm's law.
Answer the following questions:
The current flowing through the conductor is constant for a steady current flow.
Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.
Answer the following questions:
No. Ohm’s law is not universally applicable for all conducting elements.
A semiconductor diode is such an example.
Answer the following questions
3.18 (c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
Ohm's law states that: V = I x R
Hence for a low voltage V, resistance R must be very low for a high value of current.
Answer the following questions:
3.18 (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.
3.19 Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of ( ).
(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 10^{22} .
3.20 (a)
To get maximum effective resistance, combine them in series. The effective resistance will be nR.
3.20 (a)
To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.
3.20 (a)
(iii) What is the ratio of the maximum to minimum resistance?
The ratio is nR/(R/n) =
3.20 (b)
We have, equivalent resistance = 11/3
Let's break this algebraically so that we can represent it in terms of 1, 2 and 3
this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :
3.20 (b)
Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it
Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1
R = 11/5 Ω
3.20 (b)
1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.
3.20 (b)
Connect all three resistors in parallel.
Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)
R = 6/11 Ω
3.20 (c)
(a) Determine the equivalent resistance of networks shown in
Fig. 3.31.
It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,
so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is
now we have 4 such loops in series so,
Hence equivalent resistance of the circuit is 16/3 ohm.
3.20 (c)
(b) Determine the equivalent resistance of networks shown in Fig. 3.31.
It can be seen that all 5 resistors are in series, so
Equivalent Resistance = R + R + R + R + R = 5R
Hence equivalent resistance is 5R.
Answer:
First, let us find the equivalent of the infinite network,
let equivalent resistance = R'
Here from the figure, We can consider the box as a resistance of R'
Now, we can write,
equivalent resistance = R'
' =[( R')Parallel with (1)] + 1 + 1
Since resistance can never be negative we accept
, We have calculated the equivalent resistance of infinite network,
Now
Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network
= 0.5+1+1.73
=3.23 ohm
Hence current drawn from the 12V battery is 3.72 Ampere.
Given
maintained constant emf of standard cell = 1.02V, balanced point of this cell = 67.3cm
Now when the standard cell is replaced by another cell with emf = . balanced point for this cell = 82.3cm
Now as we know the relation
hence emf of another cell is 1.247V.
(b) What purpose does the high resistance of 600 k have?
If a sufficiently high current passes through galvanometer then it can get damaged. So we limit the current by adding a high resistance of 600 k .
(c) Is the balance point affected by this high resistance?
No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant current to potentiometer wire.
No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, their won't be any balance point in the wire
(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.
3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Answer:
Given,
the balance point of cell in open circuit =
value of external resistance added =
new balance point =
let the internal resistance of the cell be .
Now as we know, in a potentiometer,
hence the internal resistance of the cell will be 1.68
Chapter 3 physics class 12 ncert solutions are essential for students preparing for board exams, JEE, and NEET. This chapter covers basic concepts like electric current and resistance. Understanding these concepts is important for board exams and helps in competitive exams like JEE and NEET. Class 12 current electricity ncert solutions also forms the foundation for advanced studies in physics and electrical engineering. So, it's a key chapter for success in exams and future studies.
Where: is the angle between the normal to the area and direction of current
Relation between current density and electric field
where = conductivity and = resistivity or specific resistance of the substance
Where:
m is the mass, n is the number of electrons per unit volume, e is the charge of the electron and is the relaxation time
Grouping of Resistance
Series Grouping of resistance
Parallel Grouping of Resistance
Heat developed in a resistor
(Balanced condition)
Current Electricity Class 12 solutions for exercise questions are important from the board as well as competitive exam point of view. Following are the main topics covered in ch3 Physics Class 12.
Electric current- The concept of electric current, drift velocity, mobility etc are covered in the first part of the NCERT chapter on current electricity. All the formulas and concepts in the NCERT Class 12 Physics are important for the CBSE board exam and questions based on these parts are covered in the Current Electricity Class 12 NCERT solutions
Ohms law- Ohms law gives a linear relationship between voltage and current at a given temperature. And also discuss that not all conductors obey ohms law. There are ohmic and non-ohmic conductors. From here the term resistance is introduced and the questions related to this law is discussed in the NCERT Solutions for Class 12 Physics Chapter 3.
Resistance and Resistivity-The concepts of resistance and resistivity are discussed here. The variations of resistance and resistivity with temperature and colour code of resistance are discussed in chapter 3 Physics Class 12. Questions related to this are discussed in current electricity NCERT solutions.
Cells and their combinations-The concepts of cells, their series and parallel combinations are discussed in the CBSE Class 12 chapter 3 Current Electricity.
Kirchoff's law-This section discusses two important laws used to solve circuits namely Kirchoff's current law and voltage law. This part is important as far as competitive exams are concerned.
Wheatstone's bridge, meter bridge and potentiometer- The concepts of Wheatstone's bridge and meter bridge along with their balanced condition are discussed here. Meter bridge is an application of balanced Wheatstone's bridge. The concept of the potentiometer and its applications to compare EMFs of cells and also to find the internal resistance of cells are discussed in the chapter on current electricity. The questions from this part have high weightage for NEET and JEE main exams. 1 question from this part can be expected for JEE Main and NEET. To practice questions on this refer to chapter 3 Physics Class 12 NCERT Solutions.
As far as the CBSE board exam and competitive exams are considered, the solutions of NCERT Class 12 Physics chapter 3 are important. In NEET and JEE Main previous year papers, 7 to 10% questions are asked from this chapter. In CBSE board exam 2019, 6 marks questions are asked from current electricity. Use these CBSE NCERT solutions for Class 12 Physics chapter 3 PDF for a better score in board and competitive exams.
This NCERT Solutions for Class 12 Physics chapter 3 will help in practising questions for annual exams.
Students can solve and match their answers from this Class 12 Physics Chapter 3 NCERT solutions.
These current electricity Class 12 NCERT solutions will give proper answers that will help students in exams to score good marks.
In case of any doubt, while solving any question given in chapter 3 Physics Class 12, students can refer to this solution to know how to solve the questions and what is the answer to a question.
Comprehensive Coverage: These class 12 current electricity ncert solutions cover all the topics and questions presented in Chapter 3, ensuring a thorough understanding of "Current Electricity."
Detailed Explanations: Each solution provides step-by-step explanations, making complex concepts more accessible to students.
Clarity and Simplicity: The current electricity class 12 exercise solutions are written in clear and simple language, aiding students in better comprehension.
Practice Questions: Exercise questions are included, allowing students to practice and self-assess their understanding.
Preparation for Exams: These class 12 physics chapter 3 ncert solutions help students prepare effectively for board exams, JEE, NEET, and other competitive exams.
Foundation for Future Studies: The concepts covered in this chapter are essential for advanced topics in physics and electrical engineering.
Interconnected Concepts: It establishes connections to later chapters, emphasizing its importance as a foundational chapter.
Free Access: These solutions are available for free, making them easily accessible to all students.
These features make NCERT Class 12 Physics Chapter 3 Solutions a valuable resource for students to excel in their studies and exams.
Also, check
Excluded Content:
Certain portions of Chapter 3, "Current Electricity," have been omitted:
In CBSE board exam, around 8 to 10% questions can be expected from the chapter Current Electricity. Certain papers of CBSE ask around 15% questions from NCERT chapter 3 Current Electricity. To score well in the exam follow NCERT syllabus and the exercise given in the NCERT Book. To practice more problems can refer to NCERT exemplar.
If we analyse the previous year NEET, JEE Main and board papers we can see that the topics like meter bridge, potentiometer and problems using kcl and kvl and combinations of resistance and cells are frequently asked.
In JEE main, 2 to 3 questions can be expected from the chapter on Current Electricity. The topics like Potentiometer, meter bridge, KVL and KCL are important.
The concepts in the current electricity chapter of NCERT Class 12 Physics are integral parts of NEET exam. Along with NCERT, practicing with previous year papers and mock tests would be enough.
Current electricity carries 8% of weightage on an average for NEET exam.
Current electricity is the basics of electrical and electronics engineering-related branches. In these branches, analysis and designs of the circuit are important and the basic laws studied in the Class 12 Physics Chapter 3 NCERT solutions help for the same.
The main topics are-
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hello,
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I hope this was helpful!
Good Luck
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